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Linear Programming Problem

Linear Programming Problem. MANU SRIDHAR DEEPTHI.J VANAJA. Problem A canning company operates two canning plants. The growers are willing to supply fresh fruits in the following amounts: S1: 200 tonnes at £11/tonne S2: 310 tonnes at £10/tonne S3: 420 tonnes at £9/tonne

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Linear Programming Problem

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  1. Linear Programming Problem MANU SRIDHAR DEEPTHI.J VANAJA

  2. Problem A canning company operates two canning plants. The growers are willing to supply fresh fruits in the following amounts: S1: 200 tonnes at £11/tonne S2: 310 tonnes at £10/tonne S3: 420 tonnes at £9/tonne Shipping costs in £ per tonne are: To: Plant A Plant B From: S1 3 3.5 S2 2 2.5 S3 6 4 Plant capacities and labour costs are: Plant A Plant B Capacity 460 tonnes 560 tonnes Labour cost £26/tonne £21/tonne

  3. The canned fruits are sold at £50/tonne to the distributors. The company can sell at this price all they can produce. The objective is to find the best mixture of the quantities supplied by the three growers to the two plants so that the company maximises its profits. Formulate the problem as a linear program and explain it.

  4. Decision Variable • The amount of fruits supplied by a particular supplier to a particular plant is Xij where (i=1,2,3)and j=(1,2). • S1,S2,S3 are the growers. • The plant A is 1 and the plant B is 2.

  5. Objective Let XIJ be the number of tonnes supplied from supplier I (I = 1, 2, 3 for S1, S2, S3) to Plant J (J = 1 for A and J = 2 for B) Using the equation profit=revenue-supply costs-shipping costs-labour costs The objective is to Maximize profit. Z = 50(X11+X21+X31) + 50(X12+X22+X32) – 11(X11+X12) – 10(X21+X22) – 9(X31+X32) – 3X11-2X21-6X31-3.5X12-2.5X22-4X32 –26(X11+X21+X31) – 21(X12+X22+X32)

  6. Constraints Supply constraints: Cannot supply more than that they produce Hence, X11+X22 <= 200 X21+X22 <= 310 X31+X32 <= 420

  7. Capacity Constraint: The maximum capacity of the plant is to be taken into consideration. X11+X21+X31 <= 460 X12+X22+X32 <= 560 Non-Negativity Constraint: Xij >= 0

  8. THANK YOU

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